Do you really die if spaghettified by a black hole? Considering gravity acts by stretching spacetime and all forces (EM, nuclear etc...) are relative to spacetime, can we then consider that regardless of how stretched you are to an observer outside the black hole  as you are being spaghettified (even though I know nothing can be practically observed from outside the  black hole), you are not being spaghettified to yourself (the elevator/free fall Einstein thought experiment), and your body keeps functioning normally (making abstraction of the other falling objects that may hit you)?
 A: The equivalence principle holds in the limit of very small elevators. When far enough from a very massive object, we may have a huge but almost constant gravitational potential: strong relativistic effects, but hardly any total forces or other locally measurable effects.
Where the potential varies strongly, the equivalence principle doesn't approximately describe what happens to your body as a whole anymore. You could say that in the first case different parts of your body "want to" travel on parallel geodesics, whereas in the second case they are sufficiently divergent to overcome the forces holding the different parts together.
A: "Spaghettification" is a tidal effect caused by the different paths taken through spacetime by the particles of your body. Yes, you will notice this (and it ultimately will kill you). Your argument that electromagnetic forces are bent in the same way by spacetime isn't quite relevant, because it is not the bending of spacetime that causes spaghettification but the relative bending being different at your head and your feet, which is an objective thing agreed upon by all observers. Think of it this way: tides really exist on Earth and the atoms in the ocean really do move differently because of them. The tides near a black hole are tremendously stronger than those caused by the moon and the sun, strong enough that they will pull your head and feet apart just as the tidal forces near Earth pull the ocean on opposite sides of the Earth apart.
A: 
can we then consider that regardless of how stretched you are to an observer outside the black hole as you are being spaghettified, you are not being spaghettified to yourself

No. The curvature is an invariant fact, it is not relative.
Phenomena like length contraction lead to a geometric distortion that is stress-free. Local strain gauges or stress measurements do not detect length contraction.
That is not the case with spaghettification. With tidal forces (curvature) local strain gauges and stress measurements will detect it. When the locally-measured strains exceed your body's elastic limit your tissues will be damaged and you will die.
A: When you are being stretched by the tidal effect of gravity, you really are being stretched. You can estimate this already using Newtonian gravity, and then G.R. simply adjusts the details of the calculation. For example, consider a rod of length $L$ oriented vertically and falling towards a neutron star. The bottom of the rod is being told by gravity to accelerate towards the star at a rate
$$
\frac{G M}{(r - L/2)^2}
$$
where $M$ is the mass of the star, $r$ is the distance between the centres. Meanwhile the top of the rod is being told to accelerate towards the star at a rate
$$
\frac{G M}{(r + L/2)^2}
$$
So what does the rod do? Answer: it stretches. Its internal forces (from the bonding between the atoms) oppose this stretch and consequently a tension builds up in the rod.
To calculate this tension, you can apply Newtonian mechanics to each section of the rod, and you find the tension increases towards the centre of the rod, where it reaches the value
$$
T = \frac{1}{8} \rho A L^2 \left|\frac{dg}{dr}\right| = \frac{1}{4} \rho A L^2 \frac{GM}{r^3}
$$
where $\rho$ is the density and $A$ the cross sectional area of the rod. When this $T$ exceeds the tensile strength of the rod, the rod will break.
For example, a steel cable of length 100 metres falling towards a neutron star will be snapped by this gravitational effect when it is about two thousand kilometres from the star.
General relativity alters the details but not the main qualitative result here. So an astronaut falling towards a singularity will indeed die as their body is subjected to the extreme tidal forces that arise. As another answer already mentioned, the equivalence principle is a statement that only applies on distance and time scales small enough that the tidal effects can be ignored.
An added note on reference objects
In general relativity we are at pains to note that many statements with regard to distance and time can only be made by reference to physical objects. Can spacetime itself be regarded as such a reference object? The answer is: not in any simple sense. The basic geometric idea here is that you can think of spacetime as a 4-dimensional space (the technical term is manifold) and points in the space are called events, and there is a measure of 'spacetime distance' (called interval) between neighbouring events, which can be written:
$$
ds^2 = g_{\mu\nu} dx^{\mu} dx^{\nu}   \tag{1} 
$$
This expression gives a sum over 16 components $g_{ab}$ multiplied by
coordinate separations between the events. I won't spell out all the mathematical background. But for an understanding of spaghettification, one can note the following. What happens is that the worldlines (the trajectories in spacetime) of neighbouring parts of a physical object such as a falling rod draw apart. To be precise, let's consider an atom A and a neighbouring atom B in our rod. To begin with, the distance between these atoms might be $10^{-10}$ m or 1 angstrom, for example. That means that if you adopt the instantaneous rest frame of atom $A$, and then enquire about the spacetime location of atom $B$ at a time you would consider to be 'now', then you will find the interval $ds$ given by the above expression will have a magnitude equal to 1 angstrom. You could confirm this by comparing it with a handy ruler, made of diamond say, that you had with you. Now let some time go by as the rod falls. Maybe the rod is made of rubber, for example. Then at some later moment the expression above will give for the interval between atom $A$ and $B$ some larger number such as 2 angstroms.
Any diamond rod you had with you will stretch by much less, so it could be used
to measure this stretching of the rubber rod. If you wanted to be really accurate
then you could also calculate the expected stretching of the diamond rod and
allow for it. Or you could use a laser beam to set up a standard of distance and use that.
The point of this added note is to explain how the formula (1) applies to such experiments. In the case of a black hole the tidal stretching near the singularity will be strong enough to pull apart not just a rubber rod but also a diamond rod, or anything else.
A: As technical answers appear, I'll just address the "imagine it" part. Tidal forces near black hole are similar to centrifugal forces on a spinning body.
For a stellar mass black hole at 100 km, the tidal force over 2m (a man) is around 50,000g, so that would feel like spinning (piked, no tucking) at 7,000 RPM, or 100+ Hz.
